What is the shortest-wavelength line in the emission spectrum of the hydrogen atom?

The shortest wavelength line in the emission spectrum of the hydrogen atom is 364 nm.

Explanation:

The Balmer formula can be used to find out the wavelength which is as follow:

                      λ = b (n₂² / ( n₂² −4) )

where b = 364.56 nm and  n₂ = 3,4,5,6

It can be re-written as

ν˜ =  1 / λ

ν˜ = Rₐ ( 1/4 -1/n₂²)

Rₐ = 109,737 cm⁻¹ which is a Rydberg's constant.

For shortest wavelength, the energy would be maximum then:

ν˜ = 109737 (1/4 - 1/∞)

ν˜ = 109737 (1/4)

ν˜ =  27434 cm⁻¹

As ν˜ =  1 / λ and also λ = 1/ ν˜

λ = 1/ 27434

λ =  364 nm

the ratio of the conjugate base to the acid required to attain the desired ph.